Check sibling questions

Example 10 - Find term independent of x in (3/2 x^2 - 1/3x)^6

Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 3
Example  10 - Chapter 8 Class 11 Binomial Theorem - Part 4


Transcript

Example 10 Find the term independent of x in the expansion of (3/2 π‘₯^2 " βˆ’ " 1/3π‘₯)^6,x > 0. Calculating general term We know that general term of expansion (a + b)n is Tr + 1 = nCr (a)n–r.(b)n For general term of expansion (3/2 π‘₯^2 " βˆ’ " 1/3π‘₯)^6 Putting n = 6 , a = 3/2 π‘₯^2 , b = "βˆ’" 1/3π‘₯ Tr + 1 = 6Cr (πŸ‘/𝟐 𝒙^𝟐 )^(πŸ” βˆ’ 𝒓) ((βˆ’πŸ)/πŸ‘π’™)^𝒓 = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯^2 ")" ^((6 βˆ’ π‘Ÿ)) ((βˆ’1)/3 "Γ—" 1/π‘₯)^π‘Ÿ = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯")" ^(2(6 βˆ’ π‘Ÿ)) ((βˆ’1)/3)^π‘Ÿ (1/π‘₯)^π‘Ÿ = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) (π‘₯")" ^(12 βˆ’ 2π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯)^(βˆ’π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’ 2π‘Ÿ) (π‘₯)^(βˆ’π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’ 2π‘Ÿ βˆ’ π‘Ÿ) = 6Cr (3/2)^(6 βˆ’ π‘Ÿ) ((βˆ’1)/3)^π‘Ÿ (π‘₯")" ^(12 βˆ’3π‘Ÿ) We need to find the term independent of x So, power of x is 0 π‘₯^(12 βˆ’ 3π‘Ÿ) = x0 Comparing powers 12 – 3r = 0 12 = 3r 12/3 = r 4 = r r = 4 Putting r = 4 in (1) T4+1 = 6C4 (3/2)^(6 βˆ’ 4) ((βˆ’1)/3)^4 (π‘₯")" ^(12 βˆ’3(4)) T5 = 6C4 (3/2)^2 (1/3^4 ) (π‘₯")" ^(12 βˆ’12) = 6C4 (3^2/2^2 )(1/3^4 ) (π‘₯")" ^0 = 6C4 (1/2^2 )(3^2/3^4 ) (1) = 6C4 (1/2^2 )(1/3^2 ) = 6!/4!(6 βˆ’ 4)! (1/4)(1/9) = 6!/4!(2)! (1/4)(1/9) = (6(5)(4)!)/4!(2)! (1/4)(1/9) = (5 )/12 Hence, the term which is independent of x is 5th term = T5 = πŸ“/𝟏𝟐

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.